Special Factoring

• Feb 29th 2008, 05:06 PM
Special Factoring
\$\displaystyle (a - b)^3 - b^3\$
\$\displaystyle [(a - b) - b][(a - b)^2 + (a - b)b + b^2]\$

Is that right?
• Feb 29th 2008, 05:15 PM
Soroban

. . . Nice going!

• Feb 29th 2008, 05:50 PM
Gah, this one is driving me nuts :

Factor:

\$\displaystyle 27a^4 - a\$
\$\displaystyle a(27a^3 - 1)\$

Or can I do more?
• Feb 29th 2008, 05:55 PM
mr fantastic
Quote:

Gah, this one is driving me nuts :

Factor:

\$\displaystyle 27a^4 - a\$
\$\displaystyle a(27a^3 - 1)\$

Or can I do more?

Note that \$\displaystyle 27a^3 - 1 = (3a)^3 - 1^3\$, a difference of two perfect cubes ......
• Feb 29th 2008, 05:56 PM
I was thinking about that but didn't know what to do with the negative one. I should have known better, thanks. (Headbang)

EDIT: Just in case...

\$\displaystyle (a)(27a - 1)(27a^2 + 27a + 1)\$
• Feb 29th 2008, 06:16 PM
mr fantastic
Quote:

I was thinking about that but didn't know what to do with the negative one. I should have known better, thanks. (Headbang)

EDIT: Just in case...

\$\displaystyle (a)(27a - 1)(27a^2 + 27a + 1)\$

Should be 3a's, NOT 27a's ......
• Mar 3rd 2008, 06:17 PM
Sorry, I don't understand. I looked back over my notes but I'm kind of confused. (Thinking)
• Mar 3rd 2008, 06:36 PM
topsquark
Quote:

Gah, this one is driving me nuts :

Factor:

\$\displaystyle 27a^4 - a\$
\$\displaystyle a(27a^3 - 1)\$

Or can I do more?

\$\displaystyle 27a^3 - 1 = (3a)^3 - (1)^3\$

and
\$\displaystyle x^3 - y^3 = (x - y)(x^2 + xy + y^2)\$

So let \$\displaystyle x = 3a\$ and \$\displaystyle y = 1\$

Thus
\$\displaystyle 27a^3 - 1 = (3a)^3 - (1)^3 = (3a - 1)((3a)^2 + (3a)(1) + (1)^2)\$

\$\displaystyle = (3a - 1)(9a^2 + 3a + 1)\$

-Dan
• Mar 3rd 2008, 06:37 PM